3.28.24 \(\int \frac {\sqrt {1+x^5} (2+x^5)}{x^6 (-1-x^5+a x^{10})} \, dx\)

Optimal. Leaf size=250 \[ \frac {\left (4 \sqrt {2} a^{3/2}+\sqrt {2} \sqrt {4 a+1} \sqrt {a}+\sqrt {2} \sqrt {a}\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {x^5+1}}{\sqrt {-2 a-\sqrt {4 a+1}-1}}\right )}{5 \sqrt {4 a+1} \sqrt {-2 a-\sqrt {4 a+1}-1}}+\frac {\left (-4 \sqrt {2} a^{3/2}+\sqrt {2} \sqrt {4 a+1} \sqrt {a}-\sqrt {2} \sqrt {a}\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {x^5+1}}{\sqrt {-2 a+\sqrt {4 a+1}-1}}\right )}{5 \sqrt {4 a+1} \sqrt {-2 a+\sqrt {4 a+1}-1}}+\frac {2 \sqrt {x^5+1}}{5 x^5} \]

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Rubi [A]  time = 0.81, antiderivative size = 89, normalized size of antiderivative = 0.36, number of steps used = 18, number of rules used = 12, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {6728, 266, 47, 63, 207, 50, 6715, 824, 826, 1161, 618, 206} \begin {gather*} \frac {2}{5} \sqrt {a} \tanh ^{-1}\left (\sqrt {4 a+1}-2 \sqrt {a} \sqrt {x^5+1}\right )-\frac {2}{5} \sqrt {a} \tanh ^{-1}\left (2 \sqrt {a} \sqrt {x^5+1}+\sqrt {4 a+1}\right )+\frac {2 \sqrt {x^5+1}}{5 x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(Sqrt[1 + x^5]*(2 + x^5))/(x^6*(-1 - x^5 + a*x^10)),x]

[Out]

(2*Sqrt[1 + x^5])/(5*x^5) + (2*Sqrt[a]*ArcTanh[Sqrt[1 + 4*a] - 2*Sqrt[a]*Sqrt[1 + x^5]])/5 - (2*Sqrt[a]*ArcTan
h[Sqrt[1 + 4*a] + 2*Sqrt[a]*Sqrt[1 + x^5]])/5

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 824

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(g
*(d + e*x)^m)/(c*m), x] + Dist[1/c, Int[((d + e*x)^(m - 1)*Simp[c*d*f - a*e*g + (g*c*d - b*e*g + c*e*f)*x, x])
/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*
e^2, 0] && FractionQ[m] && GtQ[m, 0]

Rule 826

Int[((f_.) + (g_.)*(x_))/(Sqrt[(d_.) + (e_.)*(x_)]*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[2,
Subst[Int[(e*f - d*g + g*x^2)/(c*d^2 - b*d*e + a*e^2 - (2*c*d - b*e)*x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /
; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]

Rule 1161

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e - b/c, 2]},
Dist[e/(2*c), Int[1/Simp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /
; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - a*e^2, 0] && (GtQ[(2*d)/e - b/c, 0] || ( !Lt
Q[(2*d)/e - b/c, 0] && EqQ[d - e*Rt[a/c, 2], 0]))

Rule 6715

Int[(u_)*(x_)^(m_.), x_Symbol] :> Dist[1/(m + 1), Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /
; FreeQ[m, x] && NeQ[m, -1] && FunctionOfQ[x^(m + 1), u, x]

Rule 6728

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a +
b*x^n + c*x^(2*n)), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {\sqrt {1+x^5} \left (2+x^5\right )}{x^6 \left (-1-x^5+a x^{10}\right )} \, dx &=\int \left (-\frac {2 \sqrt {1+x^5}}{x^6}+\frac {\sqrt {1+x^5}}{x}-\frac {x^4 \sqrt {1+x^5} \left (-1-2 a+a x^5\right )}{-1-x^5+a x^{10}}\right ) \, dx\\ &=-\left (2 \int \frac {\sqrt {1+x^5}}{x^6} \, dx\right )+\int \frac {\sqrt {1+x^5}}{x} \, dx-\int \frac {x^4 \sqrt {1+x^5} \left (-1-2 a+a x^5\right )}{-1-x^5+a x^{10}} \, dx\\ &=\frac {1}{5} \operatorname {Subst}\left (\int \frac {\sqrt {1+x}}{x} \, dx,x,x^5\right )-\frac {1}{5} \operatorname {Subst}\left (\int \frac {\sqrt {1+x} (-1-2 a+a x)}{-1-x+a x^2} \, dx,x,x^5\right )-\frac {2}{5} \operatorname {Subst}\left (\int \frac {\sqrt {1+x}}{x^2} \, dx,x,x^5\right )\\ &=\frac {2 \sqrt {1+x^5}}{5 x^5}-\frac {\operatorname {Subst}\left (\int \frac {-2 a^2-a^2 x}{\sqrt {1+x} \left (-1-x+a x^2\right )} \, dx,x,x^5\right )}{5 a}\\ &=\frac {2 \sqrt {1+x^5}}{5 x^5}-\frac {2 \operatorname {Subst}\left (\int \frac {-a^2-a^2 x^2}{a+(-1-2 a) x^2+a x^4} \, dx,x,\sqrt {1+x^5}\right )}{5 a}\\ &=\frac {2 \sqrt {1+x^5}}{5 x^5}+\frac {1}{5} \operatorname {Subst}\left (\int \frac {1}{1-\frac {\sqrt {1+4 a} x}{\sqrt {a}}+x^2} \, dx,x,\sqrt {1+x^5}\right )+\frac {1}{5} \operatorname {Subst}\left (\int \frac {1}{1+\frac {\sqrt {1+4 a} x}{\sqrt {a}}+x^2} \, dx,x,\sqrt {1+x^5}\right )\\ &=\frac {2 \sqrt {1+x^5}}{5 x^5}-\frac {2}{5} \operatorname {Subst}\left (\int \frac {1}{\frac {1}{a}-x^2} \, dx,x,-\frac {\sqrt {1+4 a}}{\sqrt {a}}+2 \sqrt {1+x^5}\right )-\frac {2}{5} \operatorname {Subst}\left (\int \frac {1}{\frac {1}{a}-x^2} \, dx,x,\frac {\sqrt {1+4 a}}{\sqrt {a}}+2 \sqrt {1+x^5}\right )\\ &=\frac {2 \sqrt {1+x^5}}{5 x^5}+\frac {2}{5} \sqrt {a} \tanh ^{-1}\left (\sqrt {a} \left (\frac {\sqrt {1+4 a}}{\sqrt {a}}-2 \sqrt {1+x^5}\right )\right )-\frac {2}{5} \sqrt {a} \tanh ^{-1}\left (\sqrt {a} \left (\frac {\sqrt {1+4 a}}{\sqrt {a}}+2 \sqrt {1+x^5}\right )\right )\\ \end {align*}

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Mathematica [A]  time = 0.44, size = 204, normalized size = 0.82 \begin {gather*} \frac {\sqrt {4 a-2 \sqrt {4 a+1}+2} \left (4 a+\sqrt {4 a+1}+1\right ) \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {x^5+1}}{\sqrt {2 a-\sqrt {4 a+1}+1}}\right )}{10 \sqrt {a} \sqrt {4 a+1}}+\frac {\left (-4 a+\sqrt {4 a+1}-1\right ) \sqrt {2 a+\sqrt {4 a+1}+1} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {x^5+1}}{\sqrt {2 a+\sqrt {4 a+1}+1}}\right )}{5 \sqrt {2} \sqrt {a} \sqrt {4 a+1}}+\frac {2 \sqrt {x^5+1}}{5 x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[1 + x^5]*(2 + x^5))/(x^6*(-1 - x^5 + a*x^10)),x]

[Out]

(2*Sqrt[1 + x^5])/(5*x^5) + (Sqrt[2 + 4*a - 2*Sqrt[1 + 4*a]]*(1 + 4*a + Sqrt[1 + 4*a])*ArcTanh[(Sqrt[2]*Sqrt[a
]*Sqrt[1 + x^5])/Sqrt[1 + 2*a - Sqrt[1 + 4*a]]])/(10*Sqrt[a]*Sqrt[1 + 4*a]) + ((-1 - 4*a + Sqrt[1 + 4*a])*Sqrt
[1 + 2*a + Sqrt[1 + 4*a]]*ArcTanh[(Sqrt[2]*Sqrt[a]*Sqrt[1 + x^5])/Sqrt[1 + 2*a + Sqrt[1 + 4*a]]])/(5*Sqrt[2]*S
qrt[a]*Sqrt[1 + 4*a])

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IntegrateAlgebraic [A]  time = 0.54, size = 250, normalized size = 1.00 \begin {gather*} \frac {2 \sqrt {1+x^5}}{5 x^5}+\frac {\left (\sqrt {2} \sqrt {a}+4 \sqrt {2} a^{3/2}+\sqrt {2} \sqrt {a} \sqrt {1+4 a}\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {1+x^5}}{\sqrt {-1-2 a-\sqrt {1+4 a}}}\right )}{5 \sqrt {1+4 a} \sqrt {-1-2 a-\sqrt {1+4 a}}}+\frac {\left (-\sqrt {2} \sqrt {a}-4 \sqrt {2} a^{3/2}+\sqrt {2} \sqrt {a} \sqrt {1+4 a}\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {1+x^5}}{\sqrt {-1-2 a+\sqrt {1+4 a}}}\right )}{5 \sqrt {1+4 a} \sqrt {-1-2 a+\sqrt {1+4 a}}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(Sqrt[1 + x^5]*(2 + x^5))/(x^6*(-1 - x^5 + a*x^10)),x]

[Out]

(2*Sqrt[1 + x^5])/(5*x^5) + ((Sqrt[2]*Sqrt[a] + 4*Sqrt[2]*a^(3/2) + Sqrt[2]*Sqrt[a]*Sqrt[1 + 4*a])*ArcTan[(Sqr
t[2]*Sqrt[a]*Sqrt[1 + x^5])/Sqrt[-1 - 2*a - Sqrt[1 + 4*a]]])/(5*Sqrt[1 + 4*a]*Sqrt[-1 - 2*a - Sqrt[1 + 4*a]])
+ ((-(Sqrt[2]*Sqrt[a]) - 4*Sqrt[2]*a^(3/2) + Sqrt[2]*Sqrt[a]*Sqrt[1 + 4*a])*ArcTan[(Sqrt[2]*Sqrt[a]*Sqrt[1 + x
^5])/Sqrt[-1 - 2*a + Sqrt[1 + 4*a]]])/(5*Sqrt[1 + 4*a]*Sqrt[-1 - 2*a + Sqrt[1 + 4*a]])

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fricas [A]  time = 0.80, size = 103, normalized size = 0.41 \begin {gather*} \left [\frac {\sqrt {a} x^{5} \log \left (\frac {a x^{10} - 2 \, \sqrt {x^{5} + 1} \sqrt {a} x^{5} + x^{5} + 1}{a x^{10} - x^{5} - 1}\right ) + 2 \, \sqrt {x^{5} + 1}}{5 \, x^{5}}, \frac {2 \, {\left (\sqrt {-a} x^{5} \arctan \left (\frac {\sqrt {-a} x^{5}}{\sqrt {x^{5} + 1}}\right ) + \sqrt {x^{5} + 1}\right )}}{5 \, x^{5}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^5+1)^(1/2)*(x^5+2)/x^6/(a*x^10-x^5-1),x, algorithm="fricas")

[Out]

[1/5*(sqrt(a)*x^5*log((a*x^10 - 2*sqrt(x^5 + 1)*sqrt(a)*x^5 + x^5 + 1)/(a*x^10 - x^5 - 1)) + 2*sqrt(x^5 + 1))/
x^5, 2/5*(sqrt(-a)*x^5*arctan(sqrt(-a)*x^5/sqrt(x^5 + 1)) + sqrt(x^5 + 1))/x^5]

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^5+1)^(1/2)*(x^5+2)/x^6/(a*x^10-x^5-1),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:a: r
ecursive definition (in sto)  Error: Bad Argument ValueWarning, need to choose a branch for the root of a poly
nomial with parameters. This might be wrong.The choice was done assuming [a]=[-16]a: recursive definition (in
sto)  Error: Bad Argument ValueWarning, need to choose a branch for the root of a polynomial with parameters.
This might be wrong.The choice was done assuming [a]=[-4]Done

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maple [F]  time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\sqrt {x^{5}+1}\, \left (x^{5}+2\right )}{x^{6} \left (a \,x^{10}-x^{5}-1\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^5+1)^(1/2)*(x^5+2)/x^6/(a*x^10-x^5-1),x)

[Out]

int((x^5+1)^(1/2)*(x^5+2)/x^6/(a*x^10-x^5-1),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (x^{5} + 2\right )} \sqrt {x^{5} + 1}}{{\left (a x^{10} - x^{5} - 1\right )} x^{6}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^5+1)^(1/2)*(x^5+2)/x^6/(a*x^10-x^5-1),x, algorithm="maxima")

[Out]

integrate((x^5 + 2)*sqrt(x^5 + 1)/((a*x^10 - x^5 - 1)*x^6), x)

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mupad [B]  time = 2.56, size = 60, normalized size = 0.24 \begin {gather*} \frac {2\,\sqrt {x^5+1}}{5\,x^5}+\frac {\sqrt {a}\,\ln \left (\frac {a\,x^{10}+x^5-2\,\sqrt {a}\,x^5\,\sqrt {x^5+1}+1}{-4\,a\,x^{10}+4\,x^5+4}\right )}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((x^5 + 1)^(1/2)*(x^5 + 2))/(x^6*(x^5 - a*x^10 + 1)),x)

[Out]

(2*(x^5 + 1)^(1/2))/(5*x^5) + (a^(1/2)*log((a*x^10 + x^5 - 2*a^(1/2)*x^5*(x^5 + 1)^(1/2) + 1)/(4*x^5 - 4*a*x^1
0 + 4)))/5

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**5+1)**(1/2)*(x**5+2)/x**6/(a*x**10-x**5-1),x)

[Out]

Timed out

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